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hw20solutions - g x on all of R we use that it is periodic...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: March 26, 2010 Assignment 20 1. Show that if f and g are uniformly continuous on A R and if they are both bounded on A , then their product fg is uniformly continuous on A . Since f and g are both bounded on A , there exist constants M 1 and M 2 so that | f ( x ) | ≤ M 1 and | g ( x ) | ≤ M 2 on A . Fix ε > 0. Since f and g are uniformly continuous, there exist δ 1 , δ 2 > 0 so that | f ( x ) - f ( y ) | < ε/ 2 M 2 for any x, y with | x - y | < δ 1 and | g ( x ) - g ( y ) | < varepsilon/ 2 M 1 for any x, y with | x - y | < δ 2 . Then setting δ = min( δ 1 , δ 2 ), we see that for any x, y A with | x - y | < δ , | f ( x ) g ( x ) - f ( y ) g ( y ) | ≤ | f ( x ) g ( x ) - f ( y ) g ( x ) | + | f ( y ) g ( x ) - f ( y ) g ( y ) | M 2 | f ( x ) - f ( y ) | + M 2 | g ( x ) - g ( y ) | < ε as desired. 2. If f ( x ) := x and g ( x ) := sin x , show that both f and g are uniformly continuous on R , but that their product fg is not uniformly continuous on R . For any given ε > 0, one can simply choose δ = ε to show the uniform continuity of f ( x ) = x . For g ( x ), we know that sin is continuous. Thus, on the compact set [ - 2 π, 2 π ], g ( x ) is uniformly continuous. Hence, given ε > 0, there is a δ > 0 so that if | x - y | < δ and x, y [ - 2 π, π ], then | g ( x ) - g ( y ) | < ε . To get uniform continuity of
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Unformatted text preview: g ( x ) on all of R , we use that it is periodic. Thus, for any x,y ∈ R so that | x-y | < δ (assuming without loss that δ < 1) we can substract 2 πn for some n ∈ Z so that x-2 πn and y-2 πn are in [-2 π, 2 π ]. Then, | sin x-sin y | = | sin( x-2 πn )-sin( y-2 πn ) | < ε since | ( x-2 πn )-( y-2 πn ) | < δ . To show nonuniform continuity of fg on R , we shall use that sin x/x → 1 as x → and Theorem 5.4.2. Indeed, let x n = 2 πn and u n = 2 πn + 1 n . Then x n-u n → 0. x n sin x n = 0 → 0. But by the sine addition formula, u n sin u n = (2 πn + 1 /n ) sin 1 /n . But 2 πn sin 1 /n → 2 π . Thus for n sufficiently large, | fg ( x n )-fg ( u n ) | = (2 πn + 1 /n ) sin 1 /n ≥ 1 / 2 . Hence by the Nonuniform Continuity Criterion, fg is not uniformly continous. 1...
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